Question

A simple pendulum is suspended in a car moving on a circular track of radius \( R \) with a uniform speed of \( \sqrt{1.732 g R} \). The pendulum is making small oscillations in a radial direction about its equilibrium position with a time period \( T \). If the car is at rest, the time period of the pendulum is:

• A. \( T \)
• B. \( T / \sqrt{2} \)
• C. \( T \sqrt{3} \)
• D. \( 2T \)

Hint:

For a pendulum in a moving car, the effective acceleration is the vector sum of the gravitational and centripetal accelerations, leading to a change in the time period of the pendulum.

Answer 1

The Correct Option is B

When the car is moving, the pendulum experiences both the gravitational and centripetal acceleration. The effective acceleration is: \[ a_{\text{eff}} = \sqrt{g^2 + (1.732 g)^2} = 2g \] Thus, the time period of the pendulum when the car is moving is: \[ T_{\text{moving}} = \frac{T}{\sqrt{2}} \]
Hence, the time period of the pendulum when the car is moving is \( T / \sqrt{2} \).